Work out the value of $\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx$

Question

How to show that $(1)=\ln(n^2-1)?$

$$2\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx=\ln(n^2-1)\tag1$$

$n>1$

$\cos(nx)=2\cos{x}\cos[(n-1)x]-\cos[(n-2)x]$

$$2\int_{0}^{\infty}\left({\cos{x}\over x}-{{2\cos^2{x}\cos[(n-1)x]\over x}}+{\cos{x}\cos[(n-2)x]\over x}\right)\mathrm dx\tag2$$

Answer 1

\begin{align} 2\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx &= \int_0^\infty\dfrac{2\cos x-\cos(n+1)x-\cos(n-1)x}{x}\mathrm dx \\ &= \int_0^\infty\dfrac{2s}{s^2+1}-\dfrac{s}{s^2+(n+1)^2}-\dfrac{s}{s^2+(n-1)^2}\mathrm ds \\ &= \ln\dfrac{s^2+1}{\sqrt{(s^2+(n+1)^2)(s^2+(n-1)^2)}}\Big|_0^\infty \\ &= \color{blue}{\ln(n^2-1)} \end{align}

Answer 2

$\displaystyle \int\limits_0^{a\geq 0} \frac{\cos x}{x}(1-\cos(nx))dx = \int\limits_0^a \int\limits_0^n \sin(tx)dt \cos x dx = \int\limits_0^n \int\limits_0^a \sin(xt) \cos x dx dt $

$\displaystyle = \int\limits_0^n \frac{t-t\cos(a)\cos(at) - \sin(a)\sin(at) }{t^2-1} dt $

$\displaystyle =\frac{1}{2} (\ln|n^2-1|-\text{Ci}(a|1+n|)-\text{Ci}(a|1-n|)+2\text{Ci}(a))\,\, \to \,\,\frac{1}{2} \ln|n^2-1|$

for $\,\,a\to\infty\,\,$ where $\,\,|n|\neq 1\,\,$ ; $\enspace\text{Ci}(x)\,\,$ is the Cosine Integral

(e.g. http://mathworld.wolfram.com/CosineIntegral.html)

Answer 3

You may just use the complex version of Frullani's theorem, or the Laplace transform.

$$ \int_{0}^{+\infty}\frac{\cos(x)-\cos(x)\cos(nx)}{x}\,dx =\\= \int_{0}^{+\infty}\left[\frac{s}{1+s^2}-\frac{s}{2}\left(\frac{1}{(n+1)^2+s^2}+\frac{1}{(n-1)^2+s^2}\right)\right]\,ds $$ The latter is an elementary integral, equal to $\frac{1}{2}\left(\log(n-1)+\log(n+1)\right)$ for any $n>1$.

Answer 4

If you are not comfortable or familiar with the Laplace transform method suggested by @MyGlasses and @Jack D'Aurizio you could try Feynman's trick for differentiating under the integral sign. This, as I will show, is essentially the Laplace transform method in disguise.

Since $$2 \cos x \cos nx = \cos (n - 1) x + \cos (n + 1) x,$$ the integral can be rewritten as $$2 \int^\infty_0 \frac{\cos x(1 - \cos nx)}{x} \, dx = \int^\infty_0 \frac{2 \cos x - \cos (n - 1)x - \cos (n + 1) x}{x} \, dx, \quad n > 1.$$

Now consider $$I(a) = \int^\infty_0 \frac{e^{-ax} \{2 \cos x - \cos (n - 1)x - \cos (n + 1)x \}}{x} \, dx, \quad a \geqslant 0.$$ Differentiating under the integral sign with respect to $a$ we have $$I'(a) = -\int^\infty_0 \left (2 e^{-ax} \cos x - e^{-ax} \cos (n - 1)x - e^{-ax} \cos (n + 1) x \right ) \, dx.$$

Noting that $$\int^\infty_0 e^{-ax} \cos (kx) \, dx = \frac{a}{a^2 + k^2},$$ a result that can be found using integration by parts twice, one has $$I'(a) = - \left [\frac{2a}{a^2 + 1} - \frac{a}{a^2 + (n - 1)^2} - \frac{a}{a^2 + (n + 1)^2} \right ].$$ Note here that this is essentially the Laplace transform method in disguise. What we in effect have done is found the Laplace transform for the cosine function.

We require $I(0)$. Since $I(\infty) =0$, we have $$\int^\infty_0 I'(a) \, da = I(\infty) - I(0) = -I(0),$$ or $$I(0) = \int^\infty_0 \left [\frac{2a}{a^2 + 1} - \frac{a}{a^2 + (n - 1)^2} - \frac{a}{a^2 + (n + 1)^2} \right ] \, da,$$ and is exactly the same point reached by both @MyGlasses and @Jack D'Aurizio using either a method based on Laplace transforms or the complex version of Frullani's theorem.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\left.2\int_{0}^{\infty}{\cos\pars{x} \over x}\,\bracks{1 -\cos\pars{nx}}\dd x \,\right\vert_{\ n\ >\ 1} \\[5mm] = &\ \int_{0}^{\infty}{2\cos\pars{x} - \cos\pars{\bracks{n + 1}x} - \cos\pars{\bracks{n - 1}x}\over x}\,\,\dd x \\[5mm] = &\ \Re\int_{0}^{\infty}\bracks{{2\expo{\ic x} - \expo{\ic\pars{n + 1}x} - \expo{\ic\pars{n - 1}x}}} \pars{\int_{0}^{\infty}\expo{-xt}\,\dd t}\dd x \\[5mm] = &\ \Re\int_{0}^{\infty}\bracks{{2 \over t - \ic} - {1 \over t - \pars{-n - 1}\ic} - {1 \over t - \pars{-n + 1}\ic}}\dd t \\[5mm] = &\ \lim_{R \to \infty}\int_{0}^{R}\bracks{{2t \over t^{2} + 1} - {t \over t^{2} + \pars{n + 1}^{2}} - {t \over t^{2} + \pars{n - 1}^{2}}}\,\dd x \\[5mm] = &\ \lim_{R \to \infty}\braces{\ln\pars{R^{2} + 1} - {1 \over 2}\ln\pars{R^{2} + \bracks{n + 1}^{2} \over \pars{n + 1}^{2}} - {1 \over 2}\ln\pars{R^{2} + \bracks{n - 1}^{2} \over \pars{n - 1}^{2}}} \\[5mm] = &\ \bbx{\ln\pars{n^{2} - 1}\,,\qquad n > 1} \end{align}